Which graph should Laila choose? Explain your reasoning by completing the sentence. Click arrows to choose an answer from each m
enu.
1 answer:
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In Δ ABC, ∠A=120°, AB=AC=1
To draw a circumscribed circle Draw perpendicular bisectors of any of two sides.The point where these bisectors meet is the center of the circle.Mark the center as O.
Then join OA, OB, and OC.
Taking any one OA,OB and OC as radius draw the circumcircle.
Now, from O Draw OM⊥AB and ON⊥AC.
As chord AB and AC are equal,So OM and ON will also be equal.
The reason being that equal chords are equidistant from the center.
AM=MB=1/2 and AN=NC=1/2 [ perpendicular from the center to the chord bisects the chord.]
In Δ OMA and ΔONA
OM=ON [proved above]
OA is common.
MA=NA=1/2 [proved above]
ΔOMA≅ ONA [SSS]
∴ ∠OAN =∠OAM=60° [ CPCT]
In Δ OAN
OA=1
∴ OA=OB=OC=1, which is the radius of given Circumscribed circle.
Answer:
1) ΔCBF ≅ ΔCDF by (SSS)
2) ΔBFA ≅ ΔDFE by (SAS)
3) ΔCBE ≅ ΔCDA by (HL)
Step-by-step explanation:
1) Since BC ≅ DC and DF ≅ BF where CF ≅ CF (reflective property) we have;
ΔCBF ≅ ΔCDF by Side Side Side (SSS) rule of congruency
2) Since DF ≅ BF and FA ≅ FE where ∠DFE = ∠BFA (alternate angles)
Therefore;
ΔBFA ≅ ΔDFE by Side Angle Side (SAS) rule of congruency
3) Since FA ≅ FE and DF ≅ BF then where EB = FE + BF and AD = FA + DF
Where:
EB and AD are the hypotenuse sides of ΔCBE and ΔCDA respectively
We have that;
EB = AD from FE + BF = FA + DF
Where we also have BC ≅ DC
Where:
BC and DC are the legs of ΔCBE and ΔCDA respectively
Then we have the following relation;
ΔCBE ≅ ΔCDA by Hypotenuse Leg (HL).
